Or we could write this as x minus 7 squared is equal to 0.

We're asked to factor 20u squared v minus 10uv squared.

left with a squared plus 2ab plus b squared.

So this it does fit the pattern of a perfect square.

But we're done, we factored it.

And I'm gonna sum those products. So I'm taking this term.

Now, just like the last problem, we want to turn this

Polynom

polynom

Let me write down for any possible outcome of the binomial, the corresponding probability.

Sometimes there's more than one, right?

For example, instead of sending a straight

AMS Binomial Expression

n Choose m

This is the derivative, 3 over pi 5 over 1 pi.

So the easiest way to approximate it is to say, well, the simplest polynomial is just a constant, right?

This might not be completely obvious.

Normal Binomial

Displaystyle Binomial

Textstyle Binomial

That's because I've fit my parameters to the test set.

Now let's plot the following figure.

So they give us a function, f of x.

What I want to get into your brains is not the complexity of the formula. I want you to really understand how this formula is constructed.

So that's the area and it's a monomial.

Simplify the rational expression and state the domain.

So maybe, just maybe this could be a perfect square.

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So, I'll see you in the part 2 of using the quadratic equation.

But anyway I don't want to get caught up in terminology.

Minus that 3 right there.

Now, this might jump out at you immediately, as it's that special type of binomial.

The first is this term which is the cost that comes from the training set and the second is this term, which is the regularization term and what we had, we had to control the trade off between these by saying, you know, that we wanted to minimize A plus and then my regularization parameter lambda, and then times some other term B, right?

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But the important thing to recognize is this is a quadratic equation.

And what we did in this with the repeated factor is true if we went to a higher degree term.

So this checks out.

That's the same thing as 2 times 30. It equals 60. And 5 plus 12 is 17.

Let's do another one like this.

So let me just draw a little bit of a graph.

What can I do with this?

OK, now we can simplify.

We can also try adding polynomial features things like x2 square x2 square and product features x1, x2.

And it controls how big a step we take when updating my parameter theta J. And this second term here is the derivative term.